Let $D$ be a circular quadrilateral (that is a Jordan region whose boundary consists of 4 arcs
of circles all orthogonal to the unit circle) whose interior angles are all equal to 0, the vertices lie on the unit circle,
and $D$ is inside the unit disc.
Suppose also that $D$ is symmetric with respect to the real and imaginary axes, and has
one vertex $z_1=\exp(i\theta)$ where $\theta\in (0,\pi/2)$.
Then this number $\theta$ determines such a $D$ completely.

Let $f$ be the inverse of the Riemann map, so that $f$ is the conformal map from the unit disc onto $D$, $f(0)=0$ and $f'(0)>0$.

Is it true that maximum $f'(0)$ is achieved when $\theta=\pi/4$ ?

There is a strong computer evidence for this, as well as the general considerations
(where else can the maximum be?).

Alexandre, do you mean to say that $D$ is an ideal hyperbolic quadrilateral in the Poincare disk? (In other words, the arcs of $D$ are orthogonal to the unit circle). Otherwise, there is another parameter needed to define $D$, namely the angle that the arcs of $D$ make with the unit circle.
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Jeremy KahnOct 7 '12 at 1:13

Alexandre, have you looked at Chapter V of Nehari's book Conformal Mapping? He considers the general problem of finding the "Schwarz-Christoffel" mapping for hyperbolic polygons. It's complicated in general, but in your case, you should be able to write it down.
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Richard KentOct 7 '12 at 3:08

Jeremy, you are right, I consider an ideal hyperbolic quasrilateral, and I made a correction in my description.
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Alexandre EremenkoOct 7 '12 at 14:06

Richard, I know the literature, at least the classical books. This is probably the simplest problem on conformal mapping where there is no explicit solution. You can look in my preprint arXiv:1110.2696, where I wrote all I know abut this, and computed the thing numerically.
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Alexandre EremenkoOct 7 '12 at 14:10

Matti, it is interesting to know that they have a name. Do you know why Lambert? Lambert was an XVIII century mathematician who is remembered for Lambert series. Why is his name attached to these quadrilaterals? (Unless there was some other Lambert).
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Alexandre EremenkoFeb 4 '13 at 5:54